When 0.5000 grams of an unknown hydrocarbon, cxhy, is completely combusted with excess oxygen, 1.037 l co2 gas and is produced at 98.3 °c and 1.000 atm. what is the empirical formula of the hydrocarbon? (r = 0.08206 l×atm/mol×k)

C₃H₈

Explanation:

From PV = nRT

Moles of CO2 = PV / RT =(1 x 1.037)/(0.08206 x ( 273+98.3)) = 0.0340 moles of CO2

CxHy + (x + (y/2))O₂ > xCO₂ + yH₂O

there is 1 mole of C in each mole of CO2 so moles of C in CO₂ = 0.0340 moles

mass of C in CO₂ = 0.0340 x 12 = 0.41 g

This means mass of C in the hydrocarbon = 0.41g

so mass of H in the hydrocarbon = 0.50 - 0.41= 0.09 g

moles of H in the hydrocarbon = mass/molar mass = 0.09/1 = 0.09 moles

molar ratio of C:H = 0.034 : 0.09 = 1 :2.67

or 3 :8

empirical formula is C3H8

i'm wiling to try and you : )